Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int adj[8][2] = {{-1, 0},  {0, -1}, {1, 0}, {0, 1},
                       {-1, -1}, {-1, 1}, {1, 1}, {1, -1}};
const long long int LLINF = 9e18;
const int INF = 2e9;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos(-1);
const int SIZE = 100005;
int N, M, Q;
vector<int> v[SIZE], ID[SIZE];
struct table {
  int indx;
  vector<int> id;
  table() {
    indx = -1;
    id.clear();
  }
} par[20][SIZE];
int lvl[SIZE];
vector<int> merge(vector<int> &A, vector<int> &B) {
  vector<int> ans;
  for (int i = 0, j = 0;;) {
    if ((int)ans.size() == 10) return ans;
    if (i == (int)A.size() && j == (int)B.size())
      break;
    else if (i == (int)A.size() && j != (int)B.size())
      ans.push_back(B[j++]);
    else if (i != (int)A.size() && j == (int)B.size())
      ans.push_back(A[i++]);
    else {
      if (A[i] < B[j])
        ans.push_back(A[i++]);
      else if (A[i] > B[j])
        ans.push_back(B[j++]);
      else {
        ans.push_back(A[i]);
        i++;
        j++;
      }
    }
  }
  return ans;
}
void DFS(int indx, int prev) {
  lvl[indx] = lvl[prev] + 1;
  par[0][indx].indx = prev;
  par[0][indx].id = ID[indx];
  for (unsigned int i = 0; i < v[indx].size(); i++) {
    int idx = v[indx][i];
    if (idx != prev) {
      DFS(idx, indx);
    }
  }
}
void memo() {
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j <= N; j++) {
      par[i][j].id = merge(par[i - 1][j].id, par[i - 1][par[i - 1][j].indx].id);
      par[i][j].indx = par[i - 1][par[i - 1][j].indx].indx;
    }
  }
}
void LCA(int a, int b, int k) {
  if (a == b) {
    int n = min((int)ID[a].size(), k);
    printf("%d", n);
    for (int i = 0; i < n; i++) {
      printf(" %d", ID[a][i]);
    }
    printf("\n");
    return;
  }
  if (lvl[a] > lvl[b]) swap(a, b);
  int dif = lvl[b] - lvl[a];
  vector<int> A, B;
  A = par[0][a].id;
  B = par[0][b].id;
  for (int i = 0; i < 20; i++) {
    if ((dif >> i) % 2 == 1) {
      B = merge(B, par[i][b].id);
      b = par[i][b].indx;
    }
  }
  if (a == b) {
    vector<int> ans = merge(A, B);
    int n = min((int)ans.size(), k);
    printf("%d", n);
    for (int i = 0; i < n; i++) {
      printf(" %d", ans[i]);
    }
    printf("\n");
  } else {
    for (int i = 19; i >= 0; i--) {
      if (par[i][a].indx != par[i][b].indx) {
        B = merge(B, par[i][b].id);
        A = merge(A, par[i][a].id);
        a = par[i][a].indx;
        b = par[i][b].indx;
      }
    }
    vector<int> ans = merge(A, B);
    ans = merge(ans, par[0][a].id);
    ans = merge(ans, par[0][b].id);
    ans = merge(ans, par[0][par[0][a].indx].id);
    int n = min((int)ans.size(), k);
    printf("%d", n);
    for (int i = 0; i < n; i++) {
      printf(" %d", ans[i]);
    }
    printf("\n");
  }
}
int main() {
  int a, b, c;
  scanf("%d %d %d", &N, &M, &Q);
  for (int i = 1; i < N; i++) {
    scanf("%d %d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (int i = 1; i <= M; i++) {
    scanf("%d", &c);
    ID[c].push_back(i);
  }
  lvl[0] = 0;
  DFS(1, 0);
  memo();
  while (Q--) {
    scanf("%d %d %d", &a, &b, &c);
    LCA(a, b, c);
  }
  return 0;
}